As the total length is 158m, there are at most 9 water pipes with the length of 17m. It can be assumed that 9, 8, 7, 6, 5, 4, 3, 2 and 1 water pipes with the length of 17m are used, and then we can see whether the remaining length is exactly an integer multiple of 8m.
This method is to list all the possibilities of a 17 meter long water pipe, and then see which one is suitable. This method is called exhaustive method. When there are many desirable situations, this method is certainly not satisfactory, but it is still feasible in rare cases.
If 17 meter long uses X tube and 8 meter long uses Y tube, the equation can be listed.
17x+8y= 158 ( 1)
This problem needs a positive integer solution of this equation.
We use the following method to find the integer solution of this equation. First, the equation is transformed into:
8y= 158- 17x (2)
8y= 152+6- 16x-x (3)
Since 152 and 16x are multiples of 8, 6-x should also be a multiple of 8. It is only possible for x to take 6. By using the 6th generation of x, y=7 can be calculated from (2).
Therefore, there are 6 17m water pipes and 7/8m water pipes.
It can also be obtained by dividing both ends of Equation (2) by 8.
y = 158- 17x/8y = 152+6- 16x-x/8y = 19-2x+(6-x)/8
Since both x and y are integers, and 19-2x is also an integer, we know that 6-x/8 is also an integer. Obviously, 6-x/8 is an integer only when x=6. At this time, 6-x/8 = 0 and y=7.
This solution is called integer separation process or integer separation.
The bus can accommodate 54 people and the car can accommodate 36 people. At present, 378 people have to take the bus. How many cars do you need to get everyone on the bus? All the cars are just full.
Analysis: Assuming that there are X large cars and Y small cars, the equation can be obtained.
54x+36y=378
From (54,36) =18, 18|378, the original equation can be transformed into
3x+2y=2 1
And there must be an integer solution.
It is easy to see that x= 1 and y=9 are integer solutions of 3x+2y=2 1, so all integer solutions of 3x+2y=2 1 are.
x= 1+2t y=9-3t
T is any integer.
3x+2y=2 1 In addition to t=0, there are integer solutions x= 1 and y=9.
When t= 1, there are integer solutions x=3 and y=6.
When t=2, there are integer solutions x=5 and y=3.
When t=3, there are integer solutions x=7 and y=0.
Therefore, we need 1 cart and 9 carts. Or 3 carts and 6 cars; Or 5 carts and 3 cars; As long as there are seven big cars instead of small cars, 378 passengers can get on the bus and the carriages are full.
When t≤4, because y is no longer 0 and a positive integer, the solution loses its practical significance.